How to solve a biquadratic equation?
Before you start solving a biquadratic equation, you should understand how it looks and how it differs from the classical quadratic equation. Equation of the ax4+ bx2+ c = 0 is called biquadratic with one variable (an algebraic equation of the fourth degree). To reduce the equation to a quadratic form and solve it through a discriminant, it is necessary to use the variable substitution:
- ie: x2= t
And then we have a standard equation of the form at2+ bt + c = 0
The discriminant is calculated by the formula D = b2 - 4ac.
- In the case when D = 0, the equation has one single root t1= -b / 2a, and from here we get the desired solution of our equation x = sqrt (t1).
- If D> 0, the equation has two roots t1= (-b + sqrt (D)) / 2a and t2= (-b - sqrt (D)) / 2a. Do not forget about the variable entered, and we get the final solution x1,2= sqrt (t1) and x3,4= sqrt (t2)
Important note: if any of the t valuesi<0, then for D = 0 the initial biquadratic solution has no real roots, and for D> 0, there is at most one single real root.
Using the Viet theorem
Good to know: in the case when we have a reduced quadratic equation (coefficient at t2= 1), the Viet theorem is applicable, and the search for a solution is minimized:
- t1+ t2= -b
- t1* t2= c
Consider an example:
- x4- 3x2+ 2 = 0
using the substitution variable x2= t, we reduce the quadratic equation to the form t2- 3t; + 2 = 0.
- D = (-3)2- 4*1*2 = 1.
Roots of a quadratic equation t1= 2, t2= 1.
Given the introduced change of variable, we obtain the solution of the desired biquadratic equation: t1= sqrt (2); t2= -sqrt (2); t3= 1; t4= -1.
We can apply the Viet theorem to this task, since the coefficient of the variable with the highest power is 1:
- t1+ t2= 3
- t1* t2= 2
From here t1= 2, t2= 1. As we can see, the roots of the quadratic equation in both cases coincide, which means that the solution of the biquadratic equation will be the same.
In this article, we considered a special case of solving a biquadratic equation that is solved no more complicatedly than the classical quadratic equation.